12.1: Linearization

Remember linearization was the subtle art of using a nearby tangent line to approximate a function. This can be done to approximate real world functions as well.

Example \(\PageIndex{1}\)

Assume you are designing a pool that needs to be twice as long as it is wide, and exactly \(8\) feet deep. The current volume of the planned pool is \(1600\) feet\(^3\). Approximately how much more volume would you gain by extending the width by one inch (and adjusting the length accordingly)?

Solution

Clearly the volume of the pool will be length times width times depth. For the pool with top view above, we then have
\(V(x)=x\cdot 2x\cdot 8=16x^2\)

Since the pool volume was planned for \(1600\) feet\(^3\), we get
\(16x^2=1600 \longrightarrow x^2=100 \longrightarrow x=10\) feet

But the question asks for how much volume you gain by increasing \(x\) by one inch (which is \(\frac{1}{12}\) since \(x\) is measured in feet)

Instead of taking \(\displaystyle{ V\left(10+\frac{1}{12}\right)} \), we use linearization: \(V(x)\approx V(10)+V'(10)(x-10)\)

\(V(10)=\textcolor{blue}{1600}\) and \(V'(x)=32x \longrightarrow V'(10)=320\)

\(\displaystyle{ V\left(10+\frac{1}{12}\right)\approx \textcolor{blue}{1600}+320\left( \left[ 10+\frac{1}{12}\right]-10\right)=\textcolor{blue}{1600}+\textcolor{purple}{\frac{80}{3}}}\)

We estimate that the volume of the pool would increase by \(\textcolor{purple}{\displaystyle{\frac{80}{3}}}\) feet\(^3\)

Checkings our work: \(\displaystyle{ V\left(10+\frac{1}{12}\right)=16\left(10+\frac{1}{12}\right)^2=\textcolor{blue}{1600}+\textcolor{purple}{\frac{80}{3}}+\textcolor{orange}{\frac{25}{36}}}\)

Using linearization to approximate the area of the new rectangle only focuses on the linear growth or decay of the function. This can help us to determine we have an underestimate.

Example \(\PageIndex{2}\)

You are drag racing a car, which goes from 0 to 60 miles per hour in exactly 4 seconds. If you have moved a total of \(\displaystyle{\frac{1}{10}}\) mile from the starting line during these 4 seconds, approximately how far will you have gone after 6 seconds total have elapsed?

Solution

Define \(f(t)=\) distance traveled (in miles) during the first \(t\) seconds.

We are given \(\displaystyle{ f(4)=\frac{1}{10}}\), and \(f'(4)\) should be the rate of change of \(f\) (but in miles per second to match units).

\(\displaystyle{ \frac{ 60 \textit{ miles} }{1 \textit{ hour}}\textcolor{red}{\cdot \frac{ 1 \textit{ hour}}{ 60 \textit{ minutes}} \cdot \frac{ 1 \textit{ minute}}{60 \textit{ seconds}}} = \frac{1}{60} \frac{\textit{ miles}}{\textit{ second}} \longrightarrow f'(4)=\frac{1}{60}}\)

Since that's all we have to go on, we can say \(f(t)\approx f(4)+f'(4)(t-4)=\displaystyle{ \frac{1}{10}+\frac{1}{60}(t-4)}\)

So \(f(6)\approx \displaystyle{\frac{1}{10}+\frac{1}{60}(6-4)=\frac{2}{15}}\) miles

Example \(\PageIndex{3}\)

A balloon expands in the shape of a perfect sphere as air is pumped into it. If the radius is 3 inches before an additional \(\pi\) inches\(^3\) of air is pumped in, use linearization to approximate how much the radius of the balloon increases.

Hint: The volume of a sphere is \(\displaystyle{ V=\frac{4\pi}{3}r^3}\) and surface area \(\displaystyle{SA(r)=4\pi r^2}\)

Solution

Our radius should be near its start of \(3\) inches, so lets find the linearization of \(V(r)\) near \(r=3\):

\( V(r)\approx V(3)+v'(3)(r-3)\)

\(V(3)=\displaystyle{\frac{4\pi}{3}\cdot 3^3=36\pi}\)

\(\displaystyle{ V'(r)=4\pi r^2 \longrightarrow V'(3)=4\pi \cdot 3^2=36\pi}\)

Therefore, \(V(r)\approx 36\pi +36\pi(r-3)\)

Now we can set this equal to the volume increased by \(\pi\):

\(36\pi+\pi=36\pi+36\pi(r-3)\)

\(r=3+\frac{1}{36}\)

So we have the radius increase by about \(\displaystyle{\frac{1}{36}}\) inches

Isn't it interesting that for \(V(r)=\displaystyle{\frac{4\pi}{3}r^3}\) the volume of the sphere, we have \(V'(r)=4\pi r^2\) which is the formula for the surface area of a sphere.

One way to visualize this is that an infinitely small increase in the value of \(r\) will expand the volume of that sphere \(V(r)\) by an infinitely thin shell around it's surface. This thin sheet has an infinitely thin depth, but has the same area as the surface area of the sphere \(V'(r)=SA(r)\).

Many future applications in calculus will occur when we taking infinitely many things or infinitely small and/or large things. These things are all ultimately understood by taking the proper limits, which (thanks to L'Hopital's Rule) means all can relate to the derivative operation!

---